Question

Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.

${\int }_0^{2\mathrm{\pi }}{\int }_0^3{\int }_0^{\sqrt{9-r^2}}rdzdrd\theta$

Short answer

${\int }_0^{2\mathrm{\pi }}{\int }_0^3{\int }_0^{\sqrt{9-r^2}}rdzdrd\theta =18\mathrm{\pi }$

Step-by-step solution

Given

We have to find the volume by using the triple integral.

${\int }_0^{2\mathrm{\pi }}{\int }_0^3{\int }_0^{\sqrt{9-r^2}}rdzdrd\theta$

Evaluate the integral

$$\begin{gathered} V={\int }_0^{2\mathrm{\pi }}{\int }_0^3\left[{\int }_0^{\sqrt{9-r^2}}rdz\right]drd\theta \\ V={\int }_0^{2\mathrm{\pi }}\left[{\int }_0^{3}r\sqrt{9-r^2}dr\right]d\theta \end{gathered}$$

Substitute,

$$\begin{gathered} 9-r^2=t^2 \\ -2rdr=2tdt \\ dr=-\frac{t}{r}dt \\ \mathrm{When}r=0,t=\sqrt{9}=3 \\ \mathrm{When}r=3,t=9-9=0 \end{gathered}$$

Therefore,

$$\begin{gathered} V=-{\int }_0^{2\mathrm{\pi }}\left[{\int }_3^0{t}^{2}dt\right]d\theta \\ V=-{\int }_0^{2\mathrm{\pi }}{\left[\frac{t^3}{3}\right]}_3^{0}d\theta \\ V=9{\int }_0^{2\mathrm{\pi }}d\theta \\ V=9(2\mathrm{\pi }-0) \\ V=18\mathrm{\pi }\mathrm{cubic}\mathrm{units} \end{gathered}$$